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Transmitter Output Impedance
On May 5, 12:23*pm, Jim Lux wrote:
Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, ... Strangely enough, for any fixed configuration with reflections, one way to maximize radiated power is to maximize reflected power at the antenna feedpoint.:) |
Transmitter Output Impedance
On Thu, 05 May 2011 10:38:27 -0700, Jim Lux
wrote: That doesn't mean that we actually want to use a 50 ohm load in real life. Maybe I have an amplifier designed to drive a loop antenna with a resistive 10 ohm impedance. Interesting. For a community that is so tight-fisted with cash, and so brag-hearty with power claims, absolutely no one has ever tossed their hat into the ring that with lower than 50Ohm termination on their rig (and I'm not talking about 45Ohms) or with a more than 50Ohm termination on their rig (and this is certainly achievable with a 72Ohm Dipole and 70 Ohm coax) that they have then proclaimed they substantially exceeded 100W radiated (or lost to heat for that matter) by the same degree of offset from 50. Maybe that's because of tradition.. it's not only hams who do that. Well, that and the rest doesn't respond to the observation that NO ONE makes a claim of more power from a rig that DOES NOT have a source resistance of 50 Ohm. in the professional world, there's a strong tendency to make the "interfaces" between subsystems 50 ohms, even if overall system efficiency might be improved by, say, a 25 ohm impedance. Retail engineering is ruled by a market economy. You have offered a reductio-ad-absurdum of the native components offering ~10 Ohms without the cost of intervening transformers. Such an "improved efficiency" at lower cost option has been available for decades and has yet to appear as a retail product for the tight-fisted Ham market. Market logic shows by the force of economy that such a solution is not viable. a far cry from your supposition. Off by 1000%? Just an example. An incredibly poor one in the context of 100W transmitters when garden variety examples abound. I know that modern parts are MUCH lower. That deliver 97W into 50 Ohms at 30 MHZ with a nominal 12VDC supply? I smell another poor example. It might happen that you could hook up a 40 ohm load, or a 20 ohm load, and it will work just fine, and may even put out more than 100W. "May" is lazy, "Does" is more authoritative. Any reports of "Does?" The link from several days ago for the gentleman who measured the output Z being around 40 ohms would be a "does".. "Does" with how much more power? When I observe the contents of the page at that link = NOT ONE WATT MORE! In fact, if we follow the logic of lowered Rs yielding more power, his own data flips this! He reports the same 100W available at higher Rs. However, all that aside, when you observe your comment above, the link "Does not" offer any support to it. In fact on9cvd explicitly offers: "This is an ambiguous message" Which is curious when the manufacturer of the RD70HHF1 power transistors gives them a "No destroy" rating into a mismatch of 20:1. Such is the inertia of 1950s design-think with modern components. Yes, but the design limit might not be the transistor mismatch. It might be a thermal dissipation limit, or some other component that's the limiting factor. Consult the original spec. It is quite specific. The characteristic of Load VSWR tolerance: Load VSWR=20:1(All Phase): No destroy is not a dissipation reference. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On 5 mayo, 17:06, "J. C. Mc Laughlin" wrote:
Dear Wimpie: *The content of the paragraph below may well rise above the noise of this thread. I expect to learn something of value from your comments about why twice the (apparent) output Z of an amplifier was of importance and what you did to have the amplifiers conform. 73, * *Mac * N8TT -------------------"Wimpie" *wrote in message ... Hello Dave and John, snip Regarding the "academic discussion" I also agree. In my professional career where I designed several RF PA's, only 2 times the output impedance of the amplifier was of importance. *In one of these cases I couldn't meet the specs and had to insert attenuation (some waste of power…). snip With kind regards, Wim PA3DJSwww.tetech.nl Remove abc first before setting free the pigeon. -------------- J. C. Mc Laughlin Michigan U.S.A. Home: Hello, Regarding the defined output impedance. The first time was during my thesis (third harmonic peaking PA). The stability margin was not very large (expected), and it could be improved by keeping the impedance seen from the base within certain limits. My teacher said, be careful, you only have two devices (BLW76). I tried to make an exciter (based on 2SC1307 BJT) with defined output impedance, but without success. So in the end I increased the output power of the exciter and inserted a 3 dB attenuator, not elegant, but it did the job. The second time was for H-field generation where wider bandwidth was achieved by adding a second resonator. When driving from a 50 Ohms source, it had a nice Chebyshev type pass band. Because of the ripple, it shows reflection in the pass band (this happens with Chebyshev response). However when driving from a PA (that was flat within the pass band when loaded with 50 Ohms), everything went wrong. I redesigned the filter/coil combination, designed a switching PA (half bridge in class DE operation) and skipped the 50 Ohms (I designed around 8 Ohms). The PA drives the filter directly in such away that most of the time the PA sees a nice (mismatched) load (that is inductive for harmonics). This resulted in the desired pass band with significantly increased efficiency. The strength of the H-field is controlled by varying the supply voltage (PWM circuit). Leaving out the 50 Ohms in between, saved several capacitors and inductors. With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On May 5, 4:48*pm, Wimpie wrote:
On 5 mayo, 17:06, "J. C. Mc Laughlin" wrote: Dear Wimpie: *The content of the paragraph below may well rise above the noise of this thread. I expect to learn something of value from your comments about why twice the (apparent) output Z of an amplifier was of importance and what you did to have the amplifiers conform. 73, * *Mac * N8TT -------------------"Wimpie" *wrote in message .... Hello Dave and John, snip Regarding the "academic discussion" I also agree. In my professional career where I designed several RF PA's, only 2 times the output impedance of the amplifier was of importance. *In one of these cases I couldn't meet the specs and had to insert attenuation (some waste of power…). snip With kind regards, Wim PA3DJSwww.tetech.nl Remove abc first before setting free the pigeon. -------------- J. C. Mc Laughlin Michigan U.S.A. Home: Hello, Regarding the defined output impedance. The first time was during my thesis (third harmonic peaking PA). The stability margin was not very large (expected), and it could be improved by keeping the impedance seen from the base within certain limits. My teacher said, be careful, you only have two devices (BLW76). I tried to make an exciter (based on 2SC1307 BJT) with defined output impedance, but without success. *So in the end I increased the output power of the exciter and inserted a 3 dB attenuator, not elegant, but it did the job. The second time was for H-field generation where wider bandwidth was achieved by adding a second resonator. When driving from a 50 Ohms source, it had a nice Chebyshev type pass band. Because of the ripple, it shows reflection in the pass band (this happens with Chebyshev response). *However when driving from a PA (that was flat within the pass band when loaded with 50 Ohms), everything went wrong. I redesigned the filter/coil combination, designed a switching PA (half bridge in class DE operation) and skipped the 50 Ohms (I designed around 8 Ohms). The PA drives the filter directly in such away that most of the time the PA sees a nice (mismatched) load (that is inductive for harmonics). This resulted in the desired pass band with significantly increased efficiency. *The strength of the H-field is controlled by varying the supply voltage (PWM circuit). Leaving out the 50 Ohms in between, saved several capacitors and inductors. With kind regards, Wim PA3DJSwww.tetech.nl Hey Mac, are you planning on attending the Chapter 10 QCWA meeting in Cadillac Saturday? I'm planning on attending, so hope to see you there!!! Walt, W2DU |
Transmitter Output Impedance
On 5 mayo, 19:23, Jim Lux wrote:
John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJS www.tetech.nl In case of PM, tell the pigeon that abc is not in the address. |
Transmitter Output Impedance
On May 5, 5:26*pm, Wimpie wrote:
On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt |
Transmitter Output Impedance
On May 5, 9:19*pm, walt wrote:
On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g.. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt Ooops, my error, you said you're getting 130w delivered, not 120w as I said incorrectly. Walt |
Transmitter Output Impedance
On 5/5/2011 8:24 PM, walt wrote:
On May 5, 9:19 pm, wrote: On May 5, 5:26 pm, wrote: On 5 mayo, 19:23, Jim wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt Ooops, my error, you said you're getting 130w delivered, not 120w as I said incorrectly. Walt I did not present any numbers, calculations, or assumed conditions, inaccurate or otherwise. You might want to reply to Jim Lux. John - KD5YI |
Transmitter Output Impedance
On 6 mayo, 03:19, walt wrote:
On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g.. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! No he is not kidding. I observed the same. At that time I was lucky to have full access to HP, Advantest and Rohde & Schwarz equipment to double-check everything (I first blamed my own gear). You should leave the idea that all PA's have 50 Ohms output impedance, then it is easy to explain yourself. A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance. First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt The extreme cases where you can get significantly more net output power by applying mismatch, are PA's with high efficiency (class-E, - D, -DE, etc). I am currently designing a balanced class-E 500W stage. It can deliver 1 kW, but within very short time the mosfet's will explode (if the supervisory circuit doesn't act). Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On May 6, 6:30*am, Wimpie wrote:
On 6 mayo, 03:19, walt wrote: On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e..g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! No he is not kidding. I observed the same. At that time I was lucky to have full access to HP, Advantest and Rohde & Schwarz equipment to double-check everything (I first blamed my own gear). * You should leave the idea that all PA's have 50 Ohms output impedance, then it is easy to explain yourself. A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance. First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt The extreme cases where you can get significantly more net output power by applying mismatch, are PA's with high efficiency (class-E, - D, -DE, etc). * I am currently designing a balanced class-E 500W stage. It can deliver 1 kW, but within very short time the mosfet's will explode (if the supervisory circuit doesn't act). Wim PA3DJSwww.tetech.nl Wim, are you saying that by using Class E amps you are able to violate the Laws of Physics pertaining to the Conjugate Matching Theorem and the Maximum Power Delivery Theorem? I cannot agree. A load doesn't care what the source is. If the load impedance is the complex conjugate of the source, all available power will be delivered to the load. Then, if the load impedance is either increased or decreased, the power delivered will decrease. Are you now saying that the concept I just stated above is no longer true? If you are, please explain in detail why this is so. How does 'high efficiency' overcome the requirement for impedance matching in the delivery of power? And are you agreeing with an earlier poster that with a 100w source and a mismatch of 2:1 the forward power will be 135w and 15w reflected, the power delivered to the load will be 130w? If so, will you please explain in detail how this can occur? Walt |
Transmitter Output Impedance
On 6 mayo, 17:20, walt wrote:
On May 6, 6:30*am, Wimpie wrote: On 6 mayo, 03:19, walt wrote: On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience).. But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! No he is not kidding. I observed the same. At that time I was lucky to have full access to HP, Advantest and Rohde & Schwarz equipment to double-check everything (I first blamed my own gear). * You should leave the idea that all PA's have 50 Ohms output impedance, then it is easy to explain yourself. A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance. First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt The extreme cases where you can get significantly more net output power by applying mismatch, are PA's with high efficiency (class-E, - D, -DE, etc). * I am currently designing a balanced class-E 500W stage. It can deliver 1 kW, but within very short time the mosfet's will explode (if the supervisory circuit doesn't act). Wim PA3DJSwww.tetech.nl Wim, are you saying that by using Class E amps you are able to violate the Laws of Physics pertaining to the Conjugate Matching Theorem and the Maximum Power Delivery Theorem? I cannot agree. Walt, Maybe you should familiarize yourself with class E (and other high efficient topologies). Set up a simulation (or measurement) and try to apply your "conjugated match" thing. You will find out that you can't make a class E PA that operates under conjugated match, unless you are going to play with the power supply's internal resistance (so the system becomes power supply limited). My current design outputs 500W into 4.5 Ohms, however the output impedance (load pulling) is 1 Ohm. The reason for non-conjugated matched operation is that in class-E the device is in voltage saturation for about 50% of the time. During that time the device has no gain, so a device used in class-E has less gain then the same device used in a non-saturated application. In other words: tuning is designed for highest efficiency, not highest output power. A load doesn't care what the source is. If the load impedance is the complex conjugate of the source, all available power will be delivered to the load. Quote from text above: "If the load impedance is the complex conjugate of the source.....". This If-statement has a "false" result for many PAs, try to broaden your view. Then, if the load impedance is either increased or decreased, the power delivered will decrease. Are you now saying that the concept I just stated above is no longer true? If you are, please explain in detail why this is so. How does 'high efficiency' overcome the requirement for impedance matching in the delivery of power? And are you agreeing with an earlier poster that with a 100w source and a mismatch of 2:1 the forward power will be 135w and 15w reflected, the power delivered to the load will be 130w? If so, will you please explain in detail how this can occur? Yes I agree with him, see my reaction to that statement My class-E PA design that I am doing now is designed for a 4.5 Ohms nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output increases to 700W, but it will be destroyed due too non-favourite combination of Vds and Id. Walt With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On May 6, 11:43*am, Wimpie wrote:
You will find out that you can't make a class E PA that operates under conjugated match, unless you are going to play with the power supply's internal resistance ... As I understand it, Walt's approach is to pick a point inside the source at which the output becomes linear through filtering and call the impedance at that point the linear "source impedance" (including some boundary conditions). It is akin to the motion of a pendulum in a clock being linear even though the sustaining energy comes in non- linear pulses. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On May 6, 12:43*pm, Wimpie wrote:
On 6 mayo, 17:20, walt wrote: On May 6, 6:30*am, Wimpie wrote: On 6 mayo, 03:19, walt wrote: On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear.. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! No he is not kidding. I observed the same. At that time I was lucky to have full access to HP, Advantest and Rohde & Schwarz equipment to double-check everything (I first blamed my own gear). * You should leave the idea that all PA's have 50 Ohms output impedance, then it is easy to explain yourself. A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance. First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt The extreme cases where you can get significantly more net output power by applying mismatch, are PA's with high efficiency (class-E, - D, -DE, etc). * I am currently designing a balanced class-E 500W stage. It can deliver 1 kW, but within very short time the mosfet's will explode (if the supervisory circuit doesn't act). Wim PA3DJSwww.tetech.nl Wim, are you saying that by using Class E amps you are able to violate the Laws of Physics pertaining to the Conjugate Matching Theorem and the Maximum Power Delivery Theorem? I cannot agree. Walt, Maybe you should familiarize yourself with class E (and other high efficient topologies). Set up a simulation (or measurement) and try to apply your "conjugated match" thing. You will find out that you can't make a class E PA that operates under conjugated match, unless you are going to play with the power supply's internal resistance (so the system becomes power supply limited). *My current design outputs 500W into 4.5 Ohms, however the output impedance (load pulling) is 1 Ohm. The reason for non-conjugated matched operation is that in class-E the device is in voltage saturation for about 50% of the time. During that time the device has no gain, so a device used in class-E has less gain then the same device used in a non-saturated application. In other words: tuning is designed for highest efficiency, not highest output power. A load doesn't care what the source is. If the load impedance is the complex conjugate of the source, all available power will be delivered to the load. Quote from text above: "If the load impedance is the complex conjugate of the source.....". This If-statement has a "false" result for many PAs, try to broaden your view. Then, if the load impedance is either increased or decreased, the power delivered will decrease. Are you now saying that the concept I just stated above is no longer true? If you are, please explain in detail why this is so. How does 'high efficiency' overcome the requirement for impedance matching in the delivery of power? And are you agreeing with an earlier poster that with a 100w source and a mismatch of 2:1 the forward power will be 135w and 15w reflected, the power delivered to the load will be 130w? If so, will you please explain in detail how this can occur? Yes I agree with him, see my reaction to that statement My class-E PA design that I am doing now *is designed for a 4.5 Ohms nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output increases to 700W, but it will be destroyed due too non-favourite combination of Vds and Id. Walt With kind regards, Wim PA3DJSwww.tetech.nl Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. How can a 2:1 mismatch achieve 130w forward power, when that mismatch can only reflect 0.111 x 100w? Also, how can that mismatch achieve 15w of reflected power when that mismatch can only reflect 0.111 x 100w? And how can 11.111 watts of forward power add to 100w to achieve 130w? I don't see how the power reflected at a mismatched load can be affected by the nature of the source. But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5- ohm load can deliver 700w. I now understand what you mean by a mismatch can increase the power delivery. (Actually, a change in mismatch) But Wim, you pulled a fast one one us!!! Until now you didn't tell us that the source resistance of the source was 1 ohm!!! You also didn't tell us what the power would be if the load was 1 ohm (thus matching the load to the source) if the power supply was sufficiently large so the power delivery would not be limited by the power supply. So in reality, you really DON'T get an increase in power delivery by mismatching, but actually a decrease. What you're really doing is obtaining an increase in power delivery by decreasing the amount of mismatch from 4.5:1 to 2.5:1. So now I understand that you haven't violated any laws of physics, but IMHO, you have been misleading when you say you obtain a increase in power by mismatching, because that statement isn't really true, is it? take care, Walt |
Transmitter Output Impedance
On 6 mayo, 22:18, walt wrote:
On May 6, 12:43*pm, Wimpie wrote: On 6 mayo, 17:20, walt wrote: On May 6, 6:30*am, Wimpie wrote: On 6 mayo, 03:19, walt wrote: On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath.... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power.. *e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! No he is not kidding. I observed the same. At that time I was lucky to have full access to HP, Advantest and Rohde & Schwarz equipment to double-check everything (I first blamed my own gear). * You should leave the idea that all PA's have 50 Ohms output impedance, then it is easy to explain yourself. A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance. First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt The extreme cases where you can get significantly more net output power by applying mismatch, are PA's with high efficiency (class-E, - D, -DE, etc). * I am currently designing a balanced class-E 500W stage. It can deliver 1 kW, but within very short time the mosfet's will explode (if the supervisory circuit doesn't act). Wim PA3DJSwww.tetech.nl Wim, are you saying that by using Class E amps you are able to violate the Laws of Physics pertaining to the Conjugate Matching Theorem and the Maximum Power Delivery Theorem? I cannot agree. Walt, Maybe you should familiarize yourself with class E (and other high efficient topologies). Set up a simulation (or measurement) and try to apply your "conjugated match" thing. You will find out that you can't make a class E PA that operates under conjugated match, unless you are going to play with the power supply's internal resistance (so the system becomes power supply limited). *My current design outputs 500W into 4.5 Ohms, however the output impedance (load pulling) is 1 Ohm. The reason for non-conjugated matched operation is that in class-E the device is in voltage saturation for about 50% of the time. During that time the device has no gain, so a device used in class-E has less gain then the same device used in a non-saturated application. In other words: tuning is designed for highest efficiency, not highest output power. A load doesn't care what the source is. If the load impedance is the complex conjugate of the source, all available power will be delivered to the load. Quote from text above: "If the load impedance is the complex conjugate of the source.....". This If-statement has a "false" result for many PAs, try to broaden your view. Then, if the load impedance is either increased or decreased, the power delivered will decrease. Are you now saying that the concept I just stated above is no longer true? If you are, please explain in detail why this is so. How does 'high efficiency' overcome the requirement for impedance matching in the delivery of power? And are you agreeing with an earlier poster that with a 100w source and a mismatch of 2:1 the forward power will be 135w and 15w reflected, the power delivered to the load will be 130w? If so, will you please explain in detail how this can occur? Yes I agree with him, see my reaction to that statement My class-E PA design that I am doing now *is designed for a 4.5 Ohms nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output increases to 700W, but it will be destroyed due too non-favourite combination of Vds and Id. Walt With kind regards, Wim PA3DJSwww.tetech.nl Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. How can a 2:1 mismatch achieve 130w forward power, when that mismatch can only reflect 0.111 x 100w? Also, how can that mismatch achieve 15w of reflected power when that mismatch can only reflect 0.111 x 100w? And how can 11.111 watts of forward power add to 100w to achieve 130w? I don't see how the power reflected at a mismatched load can be affected by the nature of the source. But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5- ohm load can deliver 700w. I now understand what you mean by a mismatch can increase the power delivery. (Actually, a change in mismatch) But Wim, you pulled a fast one one us!!! Until now you didn't tell us that the source resistance of the source was 1 ohm!!! You also didn't tell us what the power would be if the load was 1 ohm (thus matching the load to the source) if the power supply was sufficiently large so the power delivery would not be limited by the power supply. So in reality, you really DON'T get an increase in power delivery by mismatching, but actually a decrease. What you're really doing is obtaining an increase in power delivery by decreasing the amount of mismatch from 4.5:1 to 2.5:1. So now I understand that you haven't violated any laws of physics, but IMHO, you have been misleading when you say *you obtain a increase in power by mismatching, because that statement isn't really true, is it? take care, Walt Hello Walt, Off course I agree with you that you can never have more power then the conjugated matched power, but many PAs don't operate in this regime (as my very extreme class-E case). It was the reason for mentioning: "A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance." The problem is in whether you define mismatch based on the ohmic value printed on the back of the PA, or on the actual output impedance of the PA (that you mostly don't know in case of many solid state PA's). As the actual output impedance of the PA may not be 50 Ohms (for example 100 Ohms), a 50 Ohms load (as mentioned on the back of the PA) will provide the stated power (for example 100W). By applying 100 Ohms (that is VSWR = 2 for a 50 Ohms reference), the net power will increase to 112 W. Actually it wasn't me to experience this mismatch isue first, but my father during the time that CB was very popular overhere. The above example assumes that the output impedance of the PA is independent of load. Of course this is only true when the actual voltage across the active device and current through it doesn't saturate the active device too much. I think the saturation issue is one of the advantages of a linear PA with accessible R an X tuning. You just tune for maximum output given a certain load and drive (you may use plate current as a guide also). As long as your SSB signal's PEP stays below that output power, your active device will not go into voltage saturation and IMD will likely be acceptable. In case of a wide band push-pull PA, VSWR = 2 (with inconvenient phase) may provide a load where the active devices (mosfet or BJT), will go into voltage saturation at a net output power below the rated power of the PA. For constant envelope modulation this isn't problem, but for SSB/AM it isn't good. With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Fri, 6 May 2011 13:18:16 -0700 (PDT), walt wrote:
Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. Hi Walt, Irrespective of who the author is of this statement (I've lost track), I too would like to see how this is reconciled. But seeing that you've asked several times.... 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 6, 5:50*pm, Wimpie wrote:
On 6 mayo, 22:18, walt wrote: On May 6, 12:43*pm, Wimpie wrote: On 6 mayo, 17:20, walt wrote: On May 6, 6:30*am, Wimpie wrote: On 6 mayo, 03:19, walt wrote: On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath.... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address.. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! No he is not kidding. I observed the same. At that time I was lucky to have full access to HP, Advantest and Rohde & Schwarz equipment to double-check everything (I first blamed my own gear). * You should leave the idea that all PA's have 50 Ohms output impedance, then it is easy to explain yourself. A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance. First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt The extreme cases where you can get significantly more net output power by applying mismatch, are PA's with high efficiency (class-E, - D, -DE, etc). * I am currently designing a balanced class-E 500W stage. It can deliver 1 kW, but within very short time the mosfet's will explode (if the supervisory circuit doesn't act). Wim PA3DJSwww.tetech.nl Wim, are you saying that by using Class E amps you are able to violate the Laws of Physics pertaining to the Conjugate Matching Theorem and the Maximum Power Delivery Theorem? I cannot agree. Walt, Maybe you should familiarize yourself with class E (and other high efficient topologies). Set up a simulation (or measurement) and try to apply your "conjugated match" thing. You will find out that you can't make a class E PA that operates under conjugated match, unless you are going to play with the power supply's internal resistance (so the system becomes power supply limited). *My current design outputs 500W into 4.5 Ohms, however the output impedance (load pulling) is 1 Ohm. The reason for non-conjugated matched operation is that in class-E the device is in voltage saturation for about 50% of the time. During that time the device has no gain, so a device used in class-E has less gain then the same device used in a non-saturated application. In other words: tuning is designed for highest efficiency, not highest output power. A load doesn't care what the source is. If the load impedance is the complex conjugate of the source, all available power will be delivered to the load. Quote from text above: "If the load impedance is the complex conjugate of the source.....". This If-statement has a "false" result for many PAs, try to broaden your view. Then, if the load impedance is either increased or decreased, the power delivered will decrease. Are you now saying that the concept I just stated above is no longer true? If you are, please explain in detail why this is so. How does 'high efficiency' overcome the requirement for impedance matching in the delivery of power? And are you agreeing with an earlier poster that with a 100w source and a mismatch of 2:1 the forward power will be 135w and 15w reflected, the power delivered to the load will be 130w? If so, will you please explain in detail how this can occur? Yes I agree with him, see my reaction to that statement My class-E PA design that I am doing now *is designed for a 4.5 Ohms nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output increases to 700W, but it will be destroyed due too non-favourite combination of Vds and Id. Walt With kind regards, Wim PA3DJSwww.tetech.nl Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. How can a 2:1 mismatch achieve 130w forward power, when that mismatch can only reflect 0.111 x 100w? Also, how can that mismatch achieve 15w of reflected power when that mismatch can only reflect 0.111 x 100w? And how can 11.111 watts of forward power add to 100w to achieve 130w? I don't see how the power reflected at a mismatched load can be affected by the nature of the source. But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5- ohm load can deliver 700w. I now understand what you mean by a mismatch can increase the power delivery. (Actually, a change in mismatch) But Wim, you pulled a fast one one us!!! Until now you didn't tell us that the source resistance of the source was 1 ohm!!! You also didn't tell us what the power would be if the load was 1 ohm (thus matching the load to the source) if the power supply was sufficiently large so the power delivery would not be limited by the power supply. So in reality, you really DON'T get an increase in power delivery by mismatching, but actually a decrease. What you're really doing is obtaining an increase in power delivery by decreasing the amount of mismatch from 4.5:1 to 2.5:1. So now I understand that you haven't violated any laws of physics, but IMHO, you have been misleading when you say *you obtain a increase in power by mismatching, because that statement isn't really true, is it? take care, Walt Hello Walt, Off course I agree with you that you can never have more power then the conjugated matched power, but many PAs don't operate in this regime (as my very extreme class-E case). *It was the reason for mentioning: "A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance." The problem is in whether you define mismatch based on the ohmic value printed on the back of the PA, or on the actual output impedance of the PA (that you mostly don't know in case of many solid state PA's). As the actual output impedance of the PA may not be 50 Ohms (for example 100 Ohms), a 50 Ohms load (as mentioned on the back of the PA) will provide the stated power (for example 100W). *By applying 100 Ohms (that is VSWR = 2 for a 50 Ohms reference), the net power will increase to 112 W. * Actually it wasn't me to experience this mismatch isue first, but my father during the time that CB was very popular overhere. The above example assumes that the output impedance of the PA is independent of load. *Of course this is only true when the actual voltage across the active device and current through it doesn't saturate the active device too much. I think the saturation issue is one of the advantages of a linear PA with accessible R an X tuning. You just tune for maximum output given a certain load and drive (you may use plate current as a guide also). As long as your SSB signal's PEP stays below that output power, your active device will not go into voltage saturation and IMD will likely be acceptable. In case of a wide band push-pull PA, VSWR = 2 (with inconvenient phase) may provide a load where the active devices (mosfet or BJT), will go into voltage saturation at a net output power below the rated power of the PA. For constant envelope modulation this isn't problem, but for SSB/AM it isn't good. With kind regards, Wim PA3DJSwww.tetech.nl Thanks for the reply, Wim, but you seem to be evading my questions and concerns by what appears to be going off on tangents unrelated to my questions. Perhaps I need to refresh Electronics 101, because I simply can't understand the example you provided. You have a 100-ohm source terminated with a 50-ohm load that provides 100w. But you say with the VSWR = 2 the net power will increase to 112w. How does this happen? Wim, you are making statements that seem to disagree with known principles, yet you give no explanation of how these statements can be justified in relation to the known principles. And please tell me, Wim, what RF amps do you know of that have the source impedance indicated on the back? And you refer to solid-state amps, when you know my discussion (and experience) is with tube amps with pi-network outputs. I am totally ignorant on the operation of solid-state amps. I would like to be learning something from my discussions with you, but I'm sorry, Wim, you're making me feel more stupid the more we continue. Walt |
Transmitter Output Impedance
On 5/6/2011 5:04 PM, Richard Clark wrote:
On Fri, 6 May 2011 13:18:16 -0700 (PDT), wrote: Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. Hi Walt, Irrespective of who the author is of this statement (I've lost track), I too would like to see how this is reconciled. But seeing that you've asked several times.... 73's Richard Clark, KB7QHC Here is the header from that message: From: Jim Lux Newsgroups: rec.radio.amateur.antenna Subject: Transmitter Output Impedance Date: Thu, 05 May 2011 10:23:10 -0700 Organization: JPL Information Services, InterNetNews Lines: 41 Message-ID: References: NNTP-Posting-Host: seraphic.jpl.nasa.gov Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit X-Trace: news.jpl.nasa.gov 1304616136 24458 137.79.6.96 (5 May 2011 17:22:16 GMT) X-Complaints-To: NNTP-Posting-Date: Thu, 5 May 2011 17:22:16 +0000 (UTC) User-Agent: Thunderbird 2.0.0.24 (Windows/20100228) In-Reply-To: |
Transmitter Output Impedance
On May 6, 5:24*pm, walt wrote:
You have a 100-ohm source terminated with a 50-ohm load that provides 100w. But you say with the VSWR = 2 the net power will increase to 112w. How does this happen? Walt, here is probably what Wim means by that. Source------1/2WL 50 ohm lossless------RLoad Vsource = 212v, Rsource=100 ohms If Rload = 50 ohms, PLoad = 100w If Rload = 100 ohms, PLoad=112.5w Wim must be assuming a 50 ohm SWR in both cases. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
Dear Wimpie: Many thanks. I am always on the look-out for applied
electronic problems for a colleague who is particularly skilled at crafting practice problems for the professional engineering exam (chartered engineer in Europe). Your comments about filter response reminds me of the (analog) FM receivers that had wonderful suppression of the next channel over, but horrible performance in the presence of noise from ignition - while other receivers had great performance in both respects. As you know, it had to do with phase response. Chebyshev is not always desirable. Warm regards, Mac N8TT "Wimpie" wrote in message ... On 5 mayo, 17:06, "J. C. Mc Laughlin" wrote: Dear Wimpie: The content of the paragraph below may well rise above the noise of this thread. I expect to learn something of value from your comments about why twice the (apparent) output Z of an amplifier was of importance and what you did to have the amplifiers conform. 73, Mac N8TT -------------------"Wimpie" wrote in message ... Hello Dave and John, snip Regarding the "academic discussion" I also agree. In my professional career where I designed several RF PA's, only 2 times the output impedance of the amplifier was of importance. In one of these cases I couldn't meet the specs and had to insert attenuation (some waste of power…). snip With kind regards, Wim PA3DJSwww.tetech.nl Remove abc first before setting free the pigeon. -------------- J. C. Mc Laughlin Michigan U.S.A. Home: Hello, Regarding the defined output impedance. The first time was during my thesis (third harmonic peaking PA). The stability margin was not very large (expected), and it could be improved by keeping the impedance seen from the base within certain limits. My teacher said, be careful, you only have two devices (BLW76). I tried to make an exciter (based on 2SC1307 BJT) with defined output impedance, but without success. So in the end I increased the output power of the exciter and inserted a 3 dB attenuator, not elegant, but it did the job. The second time was for H-field generation where wider bandwidth was achieved by adding a second resonator. When driving from a 50 Ohms source, it had a nice Chebyshev type pass band. Because of the ripple, it shows reflection in the pass band (this happens with Chebyshev response). However when driving from a PA (that was flat within the pass band when loaded with 50 Ohms), everything went wrong. I redesigned the filter/coil combination, designed a switching PA (half bridge in class DE operation) and skipped the 50 Ohms (I designed around 8 Ohms). The PA drives the filter directly in such away that most of the time the PA sees a nice (mismatched) load (that is inductive for harmonics). This resulted in the desired pass band with significantly increased efficiency. The strength of the H-field is controlled by varying the supply voltage (PWM circuit). Leaving out the 50 Ohms in between, saved several capacitors and inductors. With kind regards, Wim PA3DJS www.tetech.nl J. C. Mc Laughlin Michigan U.S.A. Home: |
Transmitter Output Impedance
On 7 mayo, 00:24, walt wrote:
On May 6, 5:50*pm, Wimpie wrote: On 6 mayo, 22:18, walt wrote: On May 6, 12:43*pm, Wimpie wrote: On 6 mayo, 17:20, walt wrote: On May 6, 6:30*am, Wimpie wrote: On 6 mayo, 03:19, walt wrote: On May 5, 5:26*pm, Wimpie wrote: On 5 mayo, 19:23, Jim Lux wrote: John KD5YI wrote: On 5/4/2011 7:54 PM, Jim Lux wrote: John KD5YI wrote: Acceptable is what the manufacturer recommends for his gear. What does this have to do with the device's output impedance? Absolutely nothing.. which is the point. Are we arguing the same point? but the summary is, That it is a bag of worms? I'm waiting with baited breath... Exactly.. In fact, as interesting as it would be to measure the output impedance of my radio, I started to think about what it would buy me, and came to the conclusion, almost nothing (other than satisfying curiosity). It *might* be interesting to look at (and write an article for QST/QEX or something) "optimizing radiated power". *Answering the question: do you really want a 50 ohm match on your antenna analyzer, or do you want maximum net power at the antenna feedpoint, and what that might mean for typical 100W solid state rigs, antennas, etc. (as a practical matter, this is what automatic antenna tuners actually adjust for: either minimum reflected power, or maximum fwd-ref) but it's possible that deliberately running a mismatch (as shown on your rig's SWR meter) might actually result in more radiated power. *e.g. if at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W reflected, so you're actually net 135W vs 100W; assuming your rig doesn't otherwise have any problems. Hello Jim, I agree with the PA load mismatch issue. It is possible to get (some) more net power by applying mismatch to a PA stage (from experience). But frequently it comes with too much increase in input power (so very hot heatsink). In case of high efficiency designs, the active devices may communicate to you by means of smoke or ejection of (hot) particles. Wim PA3DJSwww.tetech.nl In case of PM, tell the pigeon that abc is not in the address. Hi Wimpie and KD5YI, Will you please explain how it is possible to get more power delivered by applying a mismatch to the output of a PA? And for KD5YI specifically, I believe you have presented some inaccurate math calculations. You begin with delivering 100w into a matched load. Then you say you mismatch to 2:1 and get 135w forward and 15w reflected, leaving 120w delivered. You must be kidding!! No he is not kidding. I observed the same. At that time I was lucky to have full access to HP, Advantest and Rohde & Schwarz equipment to double-check everything (I first blamed my own gear). * You should leave the idea that all PA's have 50 Ohms output impedance, then it is easy to explain yourself. A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance. First, with a 2:1 mismatch and 100w delivered by the PA, the reflected power is 11.111w, which when added to 100w from the source, the forward power is 111.11 watts. When the reflected power that returns to the load is subtracted from the forward power, the result is 100w. You've heard the expression 'there is no free lunch'? So please explain to me, if you can, how you can deliver 120w when the source is 100w. Walt The extreme cases where you can get significantly more net output power by applying mismatch, are PA's with high efficiency (class-E, - D, -DE, etc). * I am currently designing a balanced class-E 500W stage. It can deliver 1 kW, but within very short time the mosfet's will explode (if the supervisory circuit doesn't act). Wim PA3DJSwww.tetech.nl Wim, are you saying that by using Class E amps you are able to violate the Laws of Physics pertaining to the Conjugate Matching Theorem and the Maximum Power Delivery Theorem? I cannot agree. Walt, Maybe you should familiarize yourself with class E (and other high efficient topologies). Set up a simulation (or measurement) and try to apply your "conjugated match" thing. You will find out that you can't make a class E PA that operates under conjugated match, unless you are going to play with the power supply's internal resistance (so the system becomes power supply limited). *My current design outputs 500W into 4.5 Ohms, however the output impedance (load pulling) is 1 Ohm. The reason for non-conjugated matched operation is that in class-E the device is in voltage saturation for about 50% of the time. During that time the device has no gain, so a device used in class-E has less gain then the same device used in a non-saturated application. In other words: tuning is designed for highest efficiency, not highest output power. A load doesn't care what the source is. If the load impedance is the complex conjugate of the source, all available power will be delivered to the load. Quote from text above: "If the load impedance is the complex conjugate of the source.....". This If-statement has a "false" result for many PAs, try to broaden your view. Then, if the load impedance is either increased or decreased, the power delivered will decrease. Are you now saying that the concept I just stated above is no longer true? If you are, please explain in detail why this is so. How does 'high efficiency' overcome the requirement for impedance matching in the delivery of power? And are you agreeing with an earlier poster that with a 100w source and a mismatch of 2:1 the forward power will be 135w and 15w reflected, the power delivered to the load will be 130w? If so, will you please explain in detail how this can occur? Yes I agree with him, see my reaction to that statement My class-E PA design that I am doing now *is designed for a 4.5 Ohms nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output increases to 700W, but it will be destroyed due too non-favourite combination of Vds and Id. Walt With kind regards, Wim PA3DJSwww.tetech.nl Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. How can a 2:1 mismatch achieve 130w forward power, when that mismatch can only reflect 0.111 x 100w? Also, how can that mismatch achieve 15w of reflected power when that mismatch can only reflect 0.111 x 100w? And how can 11.111 watts of forward power add to 100w to achieve 130w? I don't see how the power reflected at a mismatched load can be affected by the nature of the source. But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5- ohm load can deliver 700w. I now understand what you mean by a mismatch can increase the power delivery. (Actually, a change in mismatch) But Wim, you pulled a fast one one us!!! Until now you didn't tell us that the source resistance of the source was 1 ohm!!! You also didn't tell us what the power would be if the load was 1 ohm (thus matching the load to the source) if the power supply was sufficiently large so the power delivery would not be limited by the power supply. So in reality, you really DON'T get an increase in power delivery by mismatching, but actually a decrease. What you're really doing is obtaining an increase in power delivery by decreasing the amount of mismatch from 4.5:1 to 2.5:1. So now I understand that you haven't violated any laws of physics, but IMHO, you have been misleading when you say *you obtain a increase in power by mismatching, because that statement isn't really true, is it? take care, Walt Hello Walt, Off course I agree with you that you can never have more power then the conjugated matched power, but many PAs don't operate in this regime (as my very extreme class-E case). *It was the reason for mentioning: "A certain load that has mismatch referenced to 50 Ohms may have a nice match to a system with non-50 Ohms output impedance." The problem is in whether you define mismatch based on the ohmic value printed on the back of the PA, or on the actual output impedance of the PA (that you mostly don't know in case of many solid state PA's). As the actual output impedance of the PA may not be 50 Ohms (for example 100 Ohms), a 50 Ohms load (as mentioned on the back of the PA) will provide the stated power (for example 100W). *By applying 100 Ohms (that is VSWR = 2 for a 50 Ohms reference), the net power will increase to 112 W. * Actually it wasn't me to experience this mismatch isue first, but my father during the time that CB was very popular overhere. The above example assumes that the output impedance of the PA is independent of load. *Of course this is only true when the actual voltage across the active device and current through it doesn't saturate the active device too much. I think the saturation issue is one of the advantages of a linear PA with accessible R an X tuning. You just tune for maximum output given a certain load and drive (you may use plate current as a guide also). As long as your SSB signal's PEP stays below that output power, your active device will not go into voltage saturation and IMD will likely be acceptable. In case of a wide band push-pull PA, VSWR = 2 (with inconvenient phase) may provide a load where the active devices (mosfet or BJT), will go into voltage saturation at a net output power below the rated power of the PA. For constant envelope modulation this isn't problem, but for SSB/AM it isn't good. With kind regards, Wim PA3DJSwww.tetech.nl Thanks for the reply, Wim, but you seem to be evading my questions and concerns by what appears to be going off on tangents unrelated to my questions. Perhaps I need to refresh Electronics 101, because I simply can't understand the example you provided. You have a 100-ohm source terminated with a 50-ohm load that provides 100w. But you say with the VSWR = 2 the net power will increase to 112w. How does this happen? Wim, you are making statements that seem to disagree with known principles, yet you give no explanation of how these statements can be justified in relation to the known principles. And please tell me, Wim, what RF amps do you know of that have the source impedance indicated on the back? And you refer to solid-state amps, when you know my discussion (and experience) is with tube amps with pi-network outputs. I am totally ignorant on the operation of solid-state amps. I would like to be learning something from my discussions with you, but I'm sorry, Wim, you're making me feel more stupid the more we continue. Walt Hello Walt, The source delivers the stated power in a 50 Ohms load (so the back of the PA mentions 100W into 50 ohms, nothing more). It is the same as for my class E thing, it mentions 500W into 4.5 Ohms load. It doesn't state that the output impedance is 1 Ohms. So if one provides mismatch (referenced to 4.5 Ohms), the PA may give more net output. I hope this makes all clear. Wim |
Transmitter Output Impedance
On 7 mayo, 01:51, Cecil Moore wrote:
On May 6, 5:24*pm, walt wrote: You have a 100-ohm source terminated with a 50-ohm load that provides 100w. But you say with the VSWR = 2 the net power will increase to 112w. How does this happen? Walt, here is probably what Wim means by that. Source------1/2WL 50 ohm lossless------RLoad Vsource = 212v, Rsource=100 ohms If Rload = 50 ohms, PLoad = 100w If Rload = 100 ohms, PLoad=112.5w Wim must be assuming a 50 ohm SWR in both cases. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Hello, I assumed a black box with a voltage source in it with EMF = 213Vrms and a 100 Ohms resistor in series. Above the UHF socket it mentions: "100W into 50 Ohms". Please calculate for yourself the net power delivered into a 50 Ohms load, and into a 100 Ohms load (VSWR=2, referenced to 50 Ohms). As 50 Ohms is mentioned above the UHF socket, VSWR is referenced to 50 Ohms. Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On 7 mayo, 03:16, "J. C. Mc Laughlin" wrote:
Dear Wimpie: *Many thanks. *I am always on the look-out for applied electronic problems for a colleague who is particularly skilled at crafting practice problems for the professional engineering exam (chartered engineer in Europe). Your comments about filter response reminds me of the (analog) FM receivers that had wonderful suppression of the next channel over, but horrible performance in the presence of noise from ignition - while other receivers had great performance in both respects. *As you know, it had to do with phase response. Chebyshev is not always desirable. Warm regards, * *Mac * N8TT "Wimpie" *wrote in message ... On 5 mayo, 17:06, "J. C. Mc Laughlin" wrote: Dear Wimpie: *The content of the paragraph below may well rise above the noise of this thread. I expect to learn something of value from your comments about why twice the (apparent) output Z of an amplifier was of importance and what you did to have the amplifiers conform. 73, * *Mac * N8TT -------------------"Wimpie" *wrote in message .... Hello Dave and John, snip Regarding the "academic discussion" I also agree. In my professional career where I designed several RF PA's, only 2 times the output impedance of the amplifier was of importance. *In one of these cases I couldn't meet the specs and had to insert attenuation (some waste of power…). snip With kind regards, Wim PA3DJSwww.tetech.nl Remove abc first before setting free the pigeon. -------------- J. C. Mc Laughlin Michigan U.S.A. Home: Hello, Regarding the defined output impedance. The first time was during my thesis (third harmonic peaking PA). The stability margin was not very large (expected), and it could be improved by keeping the impedance seen from the base within certain limits. My teacher said, be careful, you only have two devices (BLW76). I tried to make an exciter (based on 2SC1307 BJT) with defined output impedance, but without success. *So in the end I increased the output power of the exciter and inserted a 3 dB attenuator, not elegant, but it did the job. The second time was for H-field generation where wider bandwidth was achieved by adding a second resonator. When driving from a 50 Ohms source, it had a nice Chebyshev type pass band. Because of the ripple, it shows reflection in the pass band (this happens with Chebyshev response). *However when driving from a PA (that was flat within the pass band when loaded with 50 Ohms), everything went wrong. I redesigned the filter/coil combination, designed a switching PA (half bridge in class DE operation) and skipped the 50 Ohms (I designed around 8 Ohms). The PA drives the filter directly in such away that most of the time the PA sees a nice (mismatched) load (that is inductive for harmonics). This resulted in the desired pass band with significantly increased efficiency. *The strength of the H-field is controlled by varying the supply voltage (PWM circuit). Leaving out the 50 Ohms in between, saved several capacitors and inductors. With kind regards, Wim PA3DJSwww.tetech.nl J. C. Mc Laughlin Michigan U.S.A. Home: ***off-topic*** Hello Mac, The Chebyshev is not may favorite filter response, but it gave the required bandwidth increase (from my head more then factor 2). When using a zero ripple response I had to use more resonating circuits and the client didn't like that. Now there is one straightforward adjustment only (that can be done in the field when required). With regards to the in-band behavior, I had an almost similar problem in an FM narrow band application that another department couldn't get within specs. They used a very steep Chebyshev filter, but they couldn't meet the internal SINAD spec…. Another filter with less pass band ripple solved the problem. Two years ago similar problem in an optical receiver. The client was happy that it could be solved by optimizing the digital detection SW. With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
=
Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. How can a 2:1 mismatch achieve 130w forward power, when that mismatch can only reflect 0.111 x 100w? Also, how can that mismatch achieve 15w of reflected power when that mismatch can only reflect 0.111 x 100w? And how can 11.111 watts of forward power add to 100w to achieve 130w? I don't see how the power reflected at a mismatched load can be affected by the nature of the source. Let's take an example of a power amplifier, followed by an isolator, and then your load. But Wim, you pulled a fast one one us!!! Until now you didn't tell us that the source resistance of the source was 1 ohm!!! This whole discussion started with the observation that most solid state amateur amplifiers probably do NOT have an output impedance of 50 ohms, and that, in fact, the output impedance isn't particularly well defined (i.e. it changes over frequency and power level). And, given that, whether you achieve a "match" may not be the best overall system solution (in terms of AC power in to radiated EM wave) The problem really is that a simple small signal linear circuit theory approach (which is what Thevenin equivalents and maximum power transfer theorem are all about) breaks down, especially when you consider that actual amplifiers have operating area limits that are some combination of current, power, and voltage, and that amplifiers and power supplies have efficiencies that vary a lot depending on the operating point. Maximum Power transfer/conjugate matching applies only at the single reference plane, and is convenient for analysis, but doesn't tell much of the story when it comes to overall system performance. A related issue is noise figure performance, where a well-matched system (which would transfer maximum power from the antenna to the LNA) sometimes isn't what you want, because it's optimally transferring the noise power from the source resistance. |
Transmitter Output Impedance
On Mon, 09 May 2011 10:04:14 -0700, Jim Lux
wrote: = Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w forward and 15w reflected with a 100w source. How can a 2:1 mismatch achieve 130w forward power, when that mismatch can only reflect 0.111 x 100w? Also, how can that mismatch achieve 15w of reflected power when that mismatch can only reflect 0.111 x 100w? And how can 11.111 watts of forward power add to 100w to achieve 130w? I don't see how the power reflected at a mismatched load can be affected by the nature of the source. Let's take an example of a power amplifier, followed by an isolator, and then your load. Hi Jim, Taking that example doesn't explain the numbers in Walt's frequently asked question (which I understand by John KD5YI's header information was originally your assertion, not Wim's). This whole discussion started with the observation that most solid state amateur amplifiers probably do NOT have an output impedance of 50 ohms, and that, in fact, the output impedance isn't particularly well defined (i.e. it changes over frequency and power level). The whole discussion that stops at that point is jejune. There is SOME output impedance at SOME operating point and the rhetorical device of saying it isn't 50 Ohms isn't very informative. The problem really is that a simple small signal linear circuit theory approach (which is what Thevenin equivalents and maximum power transfer theorem are all about) breaks down, especially when you consider that actual amplifiers have operating area limits that are some combination of current, power, and voltage, and that amplifiers and power supplies have efficiencies that vary a lot depending on the operating point. Consistently across the board, ALL of Walt's papers and publications have carefully defined the operational criteria to satisfy exactly those issues and to remove them as objections from "the whole discussion started with ... output impedance." Ignoring that contribution, and refusing to argue those points, the critics then re-inject these issues to condemn Walt's conclusion. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On 7 mayo, 01:51, Cecil Moore wrote:
On May 6, 5:24*pm, walt wrote: You have a 100-ohm source terminated with a 50-ohm load that provides 100w. But you say with the VSWR = 2 the net power will increase to 112w. How does this happen? Walt, here is probably what Wim means by that. Source------1/2WL 50 ohm lossless------RLoad Vsource = 212v, Rsource=100 ohms If Rload = 50 ohms, PLoad = 100w If Rload = 100 ohms, PLoad=112.5w Wim must be assuming a 50 ohm SWR in both cases. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Hello Cecil, Your assumption is correct! To avoid confusion, I mentioned: "that is VSWR = 2 for a 50 Ohms reference", but maybe it is because of English isn't my mother tongue. I think it is not unusual to use 50 Ohms as reference when the output specification says: "100W into 50 Ohms". With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On May 9, 3:42*pm, Wimpie wrote:
On 7 mayo, 01:51, Cecil Moore wrote: On May 6, 5:24*pm, walt wrote: You have a 100-ohm source terminated with a 50-ohm load that provides 100w. But you say with the VSWR = 2 the net power will increase to 112w. How does this happen? Walt, here is probably what Wim means by that. Source------1/2WL 50 ohm lossless------RLoad Vsource = 212v, Rsource=100 ohms If Rload = 50 ohms, PLoad = 100w If Rload = 100 ohms, PLoad=112.5w Wim must be assuming a 50 ohm SWR in both cases. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Hello Cecil, Your assumption is correct! *To avoid confusion, I mentioned: "that is VSWR = 2 for a 50 Ohms reference", but maybe it is because of English isn't my mother tongue. I think it is not unusual to use 50 Ohms as reference when the output specification says: "100W into 50 Ohms". With kind regards, Wim PA3DJSwww.tetech.nl Wim, why is it that you can't seem to answer my question concerning the amount of power reflected from a 2:1 mismatch? The following is what you said in reply, totally ignoring the details in my question: The problem really is that a simple small signal linear circuit theory approach (which is what Thevenin equivalents and maximum power transfer theorem are all about) breaks down, especially when you consider that actual amplifiers have operating area limits that are some combination of current, power, and voltage, and that amplifiers and power supplies have efficiencies that vary a lot depending on the operating point, Your response above is totally evasive, and I can't understand why you would do this. There is a specific correlation between source power, forward power and reflected power, such that with a 100w source in a 50-ohm system, a 2:1 mismatch, when the reflected power is re- reflected the forward power is 111.11w, reflected power is 11.111w, and power absorbed in the load is 100w. Now please explain with rigorous physics and math, how with a 100w source and a 2:1 mismatch you can justify a forward power of 130w and a reflected power of 15w. Is the fact that there is no way these values can exist is the reason you're evading answering the question? If someone who is unfamiliar with the truth in this instance came up with these absurd numbers, and who is willing to admit a mistake, I'm willing to forgive a mistake. But if you are stonewalling, is what it appears to be, then I'm not amused. And you should be ashamed for doing it. Walt, W2DU |
Transmitter Output Impedance
On May 9, 8:13*pm, walt wrote:
There is a specific correlation between source power, forward power and reflected power, such that with a 100w source in a 50-ohm system, a 2:1 mismatch, when the reflected power is re- reflected the forward power is 111.11w, reflected power is 11.111w, and power absorbed in the load is 100w. Walt - If the load reflects 11.111... percent of the forward power, and 100% of reflected power is re-reflected by the source, then how/why in your example is the power absorbed by the load equal to the original incident power of 100W? A re-reflection for these conditions is not 100% absorbed by the load, rather 11.111% of it is reflected back to the source. Only ~88.889% of that first re-reflection is absorbed by the load. Ditto for all further re-reflections of the first reflection. Assuming no loss in the transmission line, and phase coherence at the load for all forward power, wouldn't the load power accumulate per the table below -- which shows the initial forward power absorbed by the load with the first ten re-reflections of it by the source? 88.889 watts 1.23454321 0.137170096 0.015240969 0.001693424 0.000188156 2.09061E-05 2.32287E-06 2.58094E-07 2.86769E-08 3.18629E-09 - - - - - - - - - - 90.27785937 watts, total Then the next question is what happened to the "missing" power? RF |
Transmitter Output Impedance
On May 9, 2:42*pm, Wimpie wrote:
Your assumption is correct! *To avoid confusion, I mentioned: "that is VSWR = 2 for a 50 Ohms reference", but maybe it is because of English isn't my mother tongue. It appeared to me that Walt was talking about the SWR in a matched system, i.e. Zs= Z0=Zload=100 ohms. There is rarely an actual technical disagreement between two technical heavyweights because we all studied the same textbooks. :-) Seems to me, about 95% of all technical arguments are semantic in nature. For instance, from our earlier discussion, I believe that you consider the redistribution of RF energy associated with interference between waves to be a "reflection". To my way of thinking, a "reflection" is something that only can happen to a single wave, i.e. you and I may be presuming a different definition of "reflection" which is a semantic problem. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On May 10, 5:24*am, Richard Fry wrote:
Then the next question is what happened to the "missing" power? I was taught that by convention, i.e. by definition, the missing power was never generated in the first place - always seemed like a copout to me. :-) -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On May 10, 5:24*am, Richard Fry wrote:
RF Ignore my earlier post in this thread. I used the wrong multiplier. On further study it appears that the sum of the power contained in all perfect, coherent re-reflections from the source that will be absorbed by the load is asymptotic to the value of the power absorbed from the original incident wave (lossless transmission line). |
Transmitter Output Impedance
On 5/9/2011 6:13 PM, walt wrote:
Wim, why is it that you can't seem to answer my question concerning the amount of power reflected from a 2:1 mismatch? The following is what you said in reply, totally ignoring the details in my question: The problem really is that a simple small signal linear circuit theory approach (which is what Thevenin equivalents and maximum power transfer theorem are all about) breaks down, especially when you consider that actual amplifiers have operating area limits that are some combination of current, power, and voltage, and that amplifiers and power supplies have efficiencies that vary a lot depending on the operating point, Your response above is totally evasive, and I can't understand why you would do this. There is a specific correlation between source power, You're beating up on Wim, who didn't write that, I did. I don't have time right now, but I will respond in due course. 73 Jim, W6RMK |
Transmitter Output Impedance
On May 10, 10:01*am, Richard Fry wrote:
On further study it appears that the sum of the power contained in all perfect, coherent re-reflections from the source that will be absorbed by the load is asymptotic to the value of the power absorbed from the original incident wave (lossless transmission line). There are two mechanisms responsible for the redistribution of incident reflected power on/at the source back toward the load. 1. The incident reflected EM wave energy can be partially re-reflected from a physical impedance discontinuity at the source and merge with the forward EM wave energy. Note that "reflection" is something that happens to a single wave. 2. The components of the reflected EM wave energy that are not directly re-reflected can undergo superposition with other coherent EM waves and, associated with destructive/constructive interference, their energies can be redistributed back toward the load. Some consider this to be a re-reflection but it is NOT since it involves superposition of two (or more) EM waves. Given that b1 = s11*a1 + s12*a2 = 0, what happens to the powers, (s11*a1)^2 and (s12*a2)^2? If that energy is not moving toward the source, it is moving toward the load and contained in (s21*a1)^2 and (s22*a2)^2. That power is NOT a reflection. That's the conservation of energy principle in action associated with destructive/constructive interference. When one calculates the source impedance based on the total redistributed energy, one makes a large error when one assumes that all of the redistributed energy is "reflected". It is possible for there to exist zero re-reflected power while all of the reflected power is redistributed back toward the load through superposition associated with destructive/constructive interference. That's exactly what happens in one of W7EL's food-for-thought#1 examples. When W7EL's source sees an infinite impedance, there is zero power dissipated in the 50 ohm source resistor. However, since we are using 50 ohm lossless coax, the Z0 of the coax is equal to the Zg of the source so there are no re-reflections at the source, according to the reflection model. Is that a contradiction as W7EL seems to suggest? Absolutely not! He is simply missing the redistribution of reflected energy back toward the load as constructive interference in balance with the destructive interference at the source resistor. In s-parameter terms, if b1 = s11*a1 + s12*a2 = 0, then all of the incident reflected power must be redistributed back toward the load - but not through re-reflection - just a necessary result of destructive interference at the source resistor. I'm sure they exist but I personally do not know any hams who understand that fact of physics well recognized from the field of optics. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On 10 mayo, 03:13, walt wrote:
On May 9, 3:42*pm, Wimpie wrote: On 7 mayo, 01:51, Cecil Moore wrote: On May 6, 5:24*pm, walt wrote: You have a 100-ohm source terminated with a 50-ohm load that provides 100w. But you say with the VSWR = 2 the net power will increase to 112w. How does this happen? Walt, here is probably what Wim means by that. Source------1/2WL 50 ohm lossless------RLoad Vsource = 212v, Rsource=100 ohms If Rload = 50 ohms, PLoad = 100w If Rload = 100 ohms, PLoad=112.5w Wim must be assuming a 50 ohm SWR in both cases. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Hello Cecil, Your assumption is correct! *To avoid confusion, I mentioned: "that is VSWR = 2 for a 50 Ohms reference", but maybe it is because of English isn't my mother tongue. I think it is not unusual to use 50 Ohms as reference when the output specification says: "100W into 50 Ohms". With kind regards, Wim PA3DJSwww.tetech.nl Wim, why is it that you can't seem to answer my question concerning the amount of power reflected from a 2:1 mismatch? The following is what you said in reply, *totally ignoring the details in my question: The problem really is that a simple small signal linear circuit theory approach (which is what Thevenin equivalents and maximum power transfer theorem are all about) breaks down, especially when you consider that actual amplifiers have operating area limits that are some combination of current, power, and voltage, and that amplifiers and power supplies have efficiencies that vary a lot depending on the operating point, Your response above is totally evasive, and I can't understand why you would do this. There is a specific correlation between source power, forward power and reflected power, such that with a 100w source in a 50-ohm system, a 2:1 mismatch, when the reflected power is re- reflected the forward power is 111.11w, reflected power is 11.111w, and power absorbed in the load is 100w. Now please explain with rigorous physics and math, how with a 100w source and a 2:1 mismatch you can justify a forward power of 130w and a reflected power of 15w. Is the fact that there is no way these values can exist is the reason you're evading answering the question? If someone who is unfamiliar with the truth in this instance came up with these absurd numbers, and who is willing to admit a mistake, I'm willing to forgive a mistake. But if you are stonewalling, is what it appears to be, then I'm not amused. And you should be ashamed for doing it. Walt, W2DU Hello Walt, I thought you would be able to do this yourself, so I only stated the net delivered power to a 100 Ohms load (that is the 111W), though the delivered power into the stated 50 ohms is 100W. The example was to show you with numbers what I experienced in practice when I was 16 years old and repeated it when I had access to good equipment (mentioned before). In addition, it is not relevant to the output impedance issue of a PA, as you can calculate the increase in power due to mismatch (referenced to 50 Ohms) by just using Ohm's Law. But here it comes: We have 111W dissipated into a 100 Ohms load (from the EMF calculation, assuming 100 Ohms Zout for the PA and EMF = 212.1 Vrms). Reference for waves is 50 Ohms (according to the text on the back of the PA: "100 Watts into 50 Ohms"). We assume an electrically very small piece of 50 Ohms coaxial cable between the socket and the 100 Ohms load. This could be your reflectometer. If you like you may insert 0.5 lambda so that the PA sees 100 Ohms. 100 Ohms equals |rc| = 0.3333 (50 Ohms reference), I assume you know how to calculate this. Hence "reflected power" = 0.333^2*"incident power" = 0.111*"incident power". "net delivered power" = "incident power" – "reflected power" = 111W. "net delivered power" = (1-0.11111)*"incident power" = 111W. "incident power"= 125W, "reflected power" = 14 W Do you realize that when using transmission line resonators (or filters) the forward and reflected power INSIDE the resonator can be very high, while the net input power can be very low? Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie
wrote: Do you realize that when using transmission line resonators (or filters) the forward and reflected power INSIDE the resonator can be very high, while the net input power can be very low? And this asked of the designer of the Telstar antenna. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On 10 mayo, 21:32, Richard Clark wrote:
On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie wrote: Do you realize that when using transmission line resonators (or filters) the forward and reflected power INSIDE the resonator can be very high, while the net input power can be very low? And this asked of the designer of the Telstar antenna. 73's Richard Clark, KB7QHC Hi Richard, It doesn't matter what somebody designed in the past. I think my example regarding net power increase due to mismatch (referenced to 50 ohms) was fully clear. Stressing on forward and reflected power (several times) may give rise to the thought that somebody can't solve such a transmission line problem himself. Note that I wasn't the one that brought up this issue (as Jim noted also), but I know the stated figures aren't strange. I use a similar example in my antenna measurement classes to show that just calculating additional loss due to mismatch may go wrong completely when Rsource or Rload don't equal the reference impedance (mostly 50 Ohms). Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Tue, 10 May 2011 13:36:13 -0700 (PDT), Wimpie
wrote: On 10 mayo, 21:32, Richard Clark wrote: On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie wrote: Do you realize that when using transmission line resonators (or filters) the forward and reflected power INSIDE the resonator can be very high, while the net input power can be very low? And this asked of the designer of the Telstar antenna. 73's Richard Clark, KB7QHC Hi Richard, It doesn't matter what somebody designed in the past. This is a very curious reply given that everyone of us here has responded to the topic with past accomplishments. So it MUST matter, otherwise you have just impeached your own credentials citing experience at age 16, in work for a thesis, and work of the (recent) past. I think my example regarding net power increase due to mismatch (referenced to 50 ohms) was fully clear. So clear that it has to presume Walt doesn't know transmission line fundamentals? Stressing on forward and reflected power (several times) may give rise to the thought that somebody can't solve such a transmission line problem himself. Note that I wasn't the one that brought up this issue (as Jim noted also), but I know the stated figures aren't strange. So, by your emphasis on forward and reflected power (several times) I presume you can expand upon your application of transmission line mechanics to offer something other than an oblique comment that nothing is strange about: Now please explain with rigorous physics and math, how with a 100w source and a 2:1 mismatch you can justify a forward power of 130w and a reflected power of 15w. Is the fact that there is no way these values can exist is the reason you're evading answering the question? If, by any chance of miscommunication, you do find this last quoted statement strange; it would be clarity itself to simply say so and we will wait for Jim to sort this out. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 10, 3:32*pm, Richard Clark wrote:
On Tue, 10 May 2011 11:54:55 -0700 (PDT), Wimpie wrote: Do you realize that when using transmission line resonators (or filters) the forward and reflected power INSIDE the resonator can be very high, while the net input power can be very low? And this asked of the designer of the Telstar antenna. 73's Richard Clark, KB7QHC This is for Richard Fry, Richard, because the output source impedance of a p-network following a tube RF power amp is non-dissipative, no reflected wave enters it to be dissipated. Instead, the reflected power is totally re-reflected back into the line. The only loss from there on is due to line attenuation. The same is true when using an antenna tuner, total re- reflection. Thus, when ignoring the line loss, with a source power of 100w and a load mismatch of 2:1, the reflected power after reaching the steady state is 11.111w, and when added to the source power, the forward power is 111.11w. Here is what I believe you missed: When the 111.11w arrives at the mismatch, 11.111w is reflected, leaving 100w dissipated in the load. Correction for Richard Clark: My designs were for the dish antennas that went on the Moon buggies, and the antennas that flew on the 1st nine weather satellites. I also designed one of the quadrifilar helix antennas now flying on the polar-orbiting NOAA weather satellites. Not Telstar. Now Wim, your math is very challenging: You state 100w delivered by the source, but at one point you also state 111w is delivered to the 100-load--at another point you state that 100w is delivered to the 100- ohm load. Which is it? You also state that voltage out of the source is 212.1v--sum ting wong here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm yields 449.86w. These power values are nowhere near the values appearing in you statements. In my calculations, with 100w the voltage across a 50-ohm line is 70.71v, and across a 100-ohm line the voltage is 100v. With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you stated, thus the power-reflection coefficient is 0.1111. This means, as I've stated continually, that with 100w delivered by the source, the power reflected at the 2:1 mismatch is 11.111w, which when added to the 100w supplied by the source, makes the forward power 111.11w. Now with 11.111w of power reflected at the mismatch, this leaves 100w delivered to the 100-ohm load, not 111w. Please tell me where the 111w came from. You also haven't told us why you calculate 130w forward and 15w reflected. Someone else may have made these calculations, but it was you I asked for an explanation, because you repeated those calculations. When using the correct physics and math in this example, how can the results be so different? To close, let me present the procedure I use to calculate the total forward power--I guess it is related to Ohm's Law: With the power-reflection coefficient as prc, the forward power PF = 1/ (1 - prc). So Wim, can you clarify the confusion you appear to have made? Walt |
Transmitter Output Impedance
On Tue, 10 May 2011 14:37:35 -0700 (PDT), walt wrote:
Correction for Richard Clark: My designs were for the dish antennas that went on the Moon buggies, and the antennas that flew on the 1st nine weather satellites. I also designed one of the quadrifilar helix antennas now flying on the polar-orbiting NOAA weather satellites. Not Telstar. Hi Walt, Your paygrade is still intact. Now Wim, your math is very challenging: You state 100w delivered by the source, but at one point you also state 111w is delivered to the 100-load--at another point you state that 100w is delivered to the 100- ohm load. Which is it? There seems to be a contagion of problems infecting this thread. Quite typical for this class of discussion, unfortunately, with so many "substitutions" offered in place of sticking with the subject. You also haven't told us why you calculate 130w forward and 15w reflected. Someone else may have made these calculations, but it was you I asked for an explanation, because you repeated those calculations. However, it is still Jim's cross to bear on this slow path to Calvary. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 10, 4:37*pm, walt wrote:
This is for Richard Fry, Richard, because the output source impedance of a p-network following a tube RF power amp is non-dissipative, no reflected wave enters it to Here is what I believe you missed: When the 111.11w arrives at the mismatch, 11.111w is reflected, leaving 100w dissipated in the load. Walt, Let us start at T=0. The transmitter is keyed, and instantaneously delivers 100W to the input of a lossless transmission line. That 100W travels to the end of the transmission line and encounters a load connected there having a 2:1 mismatch to the Zo of the transmission line. That load will dissipate about 88.889W, and reflect about 11.111W back to the source. Likewise, the load will dissipate only 88.889% of the RE-reflected power from the source for the same reason it dissipated only 88.889% of the power in the original incident wave. So even if the RE-reflection from the source is lossless, and phase- coherent with the original incident power at the load, how does this yield the 100W of load dissipation that you posit? RF |
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